Question

Solve $\int e^{2x}\left(\frac{\sin 4x-2}{1-\cos 4x}\right)dx$

• A. $\frac{1}{2}{e}^{2x}\cot 2x+c$
• B. $-\frac{1}{2}{e}^{2x}\cot 2x+c$
• C. $-2e^{2x}\cot 2x+c$
• D. $2e^{2x}\cot 2x+c$

Answer 1

The correct option is A $\frac{1}{2}{e}^{2x}\cot 2x+c$

Solution -

$\int e{x}^{2x}\left(\frac{sin4x-2}{1-cos4x}\right)dx$

Put $2x=t$ $dx=\frac{dt}{2}$

$\frac{1}{2}\int e^t\left(\frac{sin2t-2}{1-cos2t}\right)dt$

$\frac{1}{2}\int e^t\left(\frac{2sintcost-2}{1-(1-2si{n}^{2}t)}\right)dt$

$\frac{1}{2}\int e^t(cott-cose{c}^{2}t)dt$

we know that $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

$=\frac{1}{2}{e}^{t}cot2x+c$

$=\frac{1}{2}{e}^{2x}cot2x+c$

A is correct.